Seema knows the mass of a basketball. What other information is needed to find the ball’s potential energy? the volume and heigh

Answer:

Explanation:

The measurement of pressure is indicated as where p denotes the pressure, signifies density, and h represents height

Given values include pressure , gravity's acceleration , and height =1.163 m

Answer:

the temperature on the left side is 1.48 times greater than that on the right

Explanation:

GIVEN DATA:

T1 = 525 K

T2 = 275 K

It is known that

n and v are constant on both sides. Therefore we have

..............1

let the final pressure be P and the temperature

..................2

similarly

.............3

divide equation (2) by equation (3)

thus, the left side temperature equals 1.48 times the right side temperature

Answer:

Explanation:

The data indicates that point A is located midway between two charges.

To calculate the electric field at point A, we begin with the field produced by charge -Q ( 6e⁻ ) at A:

= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴

= 13.82 x 10⁻⁶ N/C

This field points towards Q⁻.

A similar field will arise from the charge Q⁺, but it will direct away from Q⁺ toward Q⁻.

To find the resultant field, we add these contributions:

= 2 x 13.82 x 10⁻⁶

= 27.64 x 10⁻⁶ N/C

For the force acting on an electron placed at A:

= charge x field

= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶

= 44.22 x 10⁻²⁵ N

a) 0.13*τ; b) 2.08*τ. To describe the discharging process of a capacitor through a resistor, consider the following: Q(t) = Qo * exp(-t/τ) to signify a loss of 1/8 of its charge. In this scenario, Q(t) = 7/8 * Qo = 7/8 * exp(-t/τ). By rearranging, we have ln(7/8)*τ = -t, thus t = -ln(7/8)*τ = 0.13. For a loss of 7/8 of its charge, we use Q(t) = 1/7 * Qo * exp(-t/τ), leading to t = -ln(1/8)*τ = 2.08.

This is somewhat misleading, and I encountered the same question in my homework. An electric field strength of 1*10^5 N/C is provided, along with a drag force of 7.25*10^-11 N, and the critical detail is that it maintains a constant velocity, indicating that the particle is in equilibrium and not accelerating.
<span>To solve, utilize F=(K*Q1*Q2)/r^2 </span>
<span>You'll want to equate F with the drag force, where the electric field strength translates to (K*Q2)/r^2; substituting the values results in </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>